Notes
Basics and counting
- $\displaystyle P(\text{event}) = \frac{\text{favorable}}{\text{total}}$ for equally likely outcomes. $P(\text{not } A) = 1 - P(A)$.
- Multiplication principle: multiply the number of ways for each stage. Permutations count ordered arrangements; combinations $_nC_r$ count unordered groups.
Compound events
- Independent: $P(A \text{ and } B) = P(A)P(B)$. Dependent (for example, without replacement): $P(A \text{ and } B) = P(A)P(B \mid A)$.
- Addition rule: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$.
Conditional probability and tables
- $\displaystyle P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$. In a two-way table, divide by the row or column total for the given group.
- Events are independent if $P(A \mid B) = P(A)$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
Six of the 36 equally likely outcomes sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
Order does not matter: $\displaystyle _8C_2 = \frac{8 \cdot 7}{2} = 28$.
For independent events, multiply: $0.6 \cdot 0.3 = 0.18$.
$0.45 + 0.35 - 0.15 = 0.65$.
$\displaystyle P(\text{band} \mid \text{Spanish}) = \frac{0.24}{0.6} = 0.4$.
$10 \cdot 9 \cdot 8 \cdot 7 = 5040$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A bag holds 4 red marbles and 6 blue marbles. Two marbles are drawn one after the other.
(a) Find the probability both are red if the first marble is NOT replaced.
Show the answer and scoring guide
$\displaystyle \frac{4}{10} \cdot \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$.
Check off the points you earned:
(b) Find the probability both are red if the first marble IS replaced.
Show the answer and scoring guide
The draws are independent: $\displaystyle \frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} = \frac{4}{25}$.
Check off the points you earned:
(c) Without replacement, find the probability that at least one marble is blue.
Show the answer and scoring guide
"At least one blue" is the complement of "both red": $\displaystyle 1 - \frac{2}{15} = \frac{13}{15}$.
Check off the points you earned: